In which situations is work being done? ( You can pick more than 1)

2 answers:

7 0

A balloon expanding would count as work because it is moving and none of the others are. However, I think that James pedaling also works because at least he is using his leg muscles even though he's not moving. He is working because he is exercising himself and that count as working. I would chose James over all of them.

Ivan3 years ago

4 0

Answer:

Andy’s car rolls down a hill with the engine off.

A balloon expands as it is heated by the Sun.

James pedals a stationary exercise cycle that doesn’t move.

Explanation:

1). Hannah leans against her locker, but it doesn’t move.

Since Hannah is at rest and there is no displacement so there is no work on Hannah by the contact force

2). Andy’s car rolls down a hill with the engine off.

Since car is rolled down the hill so here work is done on the car by force of gravity on the car

3). A wind turbine stands still on a sunny, windless day.

Since wind turbine is at rest so there will be no work on the turbine due to force of wind

4). A balloon expands as it is heated by the Sun.

Here since the volume of balloon is increased so work is done by the internal air of balloon to expand it.

5). James pedals a stationary exercise cycle that doesn’t move.

While pedaling since the force exerted by the cyclist and the pedal goes down so work is done by the cyclist to move the pedal up and down

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Which of the following has the most biomass?

Crazy boy [7]

Where are the questions so that I can deliver a more accurate answer.

8 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$